The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. an article reports the following summary data on intake for a sample of males age 65−74 years: n = 114, x = 11.2, and s = 6.58. does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (use α = 0.05.) state the appropriate null and alternative hypotheses.

Step-by-step explanation:

Hello!

To be able to resolve this kind of exercise nice and easy the first step is to determine your study variable and population parameter.

The variable is

X: daily dietary allowance for zinc for a male 50 years old or older. (mg/day)

The parameter of interest is the population mean. (μ)

You need to test if the average zinc intake is bellowed the recommended allowance, symbolically μ < 15. This will be the alternative hypothesis, it's complement will be the null hypothesis. (easy tip to detect the null hypothesis fast, it always carries the = sign)

The hypothesis is:

H₀: μ ≥ 15

H₁: μ < 15

α: 0,05

Now you have no information about the variable distribution. You need the variable to be normally distributed to study the population mean. Since the sample is large enough, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal, this way, you can use the approximate Z to make the test.

X[bar]≈N(μ;σ²/n)

Z=  X[bar]-μ ≈N(0;1)

√(σ²/n)

Z=  11.2 - 15  = -6.17

6.58/√114

The critical region of this test is one-tailed (left) the critical value is:

If Z ≤ -1.64, you reject the null hypothesis.

If Z > -1.64, you support the null hypothesis

The calculated value is less than the critical value, so the decision is to reject the null hypothesis.

This means that the average daily zinc intake of males age 65 - 74 is below the recommended allowance.

I hope it helps!

-2

7 - 9 = -2

reversed:

-2 + 9 = 7